JEE Main 2025 Jan 24 Shift 1, Mathematics Q17: Quadratic Equations

JEE Main2025Jan 24, Shift 1Mathematics

The product of all the rational roots of the equation $(x^{2} - 9 x + 11)^{2} - (x - 4) (x - 5) = 3$ , is equal to :

Solution

Expand the linear product: $(x - 4) (x - 5) = x^{2} - 9 x + 20$ . Let $t = x^{2} - 9 x$ . The equation becomes

$(t + 11)^{2} - (t + 20) = 3$

$t^{2} + 22 t + 121 - t - 20 - 3 = 0$

$t^{2} + 21 t + 98 = 0 \Rightarrow (t + 7) (t + 14) = 0$

So $x^{2} - 9 x = - 7$ or $x^{2} - 9 x = - 14$ .

Case 1: $x^{2} - 9 x + 7 = 0$ . Discriminant $= 81 - 28 = 53$ (not a perfect square) $\Rightarrow$ irrational roots.

Case 2: $x^{2} - 9 x + 14 = 0 \Rightarrow (x - 2) (x - 7) = 0 \Rightarrow x = 2, 7$ . Both rational.

Product of rational roots $= 2 \cdot 7 = 14$ .

$∴$ Option (1).

Concept behind this question

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